$\dfrac{d}{dx}[2\text{cos}(x)-3\text{ln}(x)-1]=$
Recall that ${\dfrac{d}{dx}[\text{ln}(x)]=\dfrac1x}$ and ${\dfrac{d}{dx}[\text{cos}(x)]=-\text{sin}(x)}$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[2\text{cos}(x)-3\text{ln}(x)-1] \\\\ &=2{\dfrac{d}{dx}[\text{cos}(x)]}-3{\dfrac{d}{dx}[\text{ln}(x)]}-\dfrac{d}{dx}[1] \\\\ &=2[{-\text{sin}(x)}]-3\cdot{\dfrac1x}-0 \\\\ &=-2\text{sin}(x)-\dfrac3x \end{aligned}$ In conclusion, $\dfrac{d}{dx}[2\text{cos}(x)-3\text{ln}(x)-1]=-2\text{sin}(x)-\dfrac3x$